marginal probability density function

**ben.mahoney@tesco.net** · March 7th 2009, 09:24 AM

fX,Y(x,y) = 6(1-x-y) for x and y defined over the unit square, subject to the restriction that 0<x+y<1. Find the marginal probability function for X.

I have started by integrating with limits 1-x and -x getting a final answer of 18(x-1-2x^2). Anyone else get this answer?

Thanks

**Moo** · March 8th 2009, 12:18 AM

Hello,

Originally Posted by ben.mahoney@tesco.net

fX,Y(x,y) = 6(1-x-y) for x and y defined over the unit square, subject to the restriction that 0<x+y<1. Find the marginal probability function for X.

I have started by integrating with limits 1-x and -x getting a final answer of 18(x-1-2x^2). Anyone else get this answer?

Thanks

Since you integrate over the unit square, $x \in [0,1]$ and $y \in [0,1]$
so the integral over y should go from 0 to 1-x.

And this gives something like $3(1-x)^2$

**ben.mahoney@tesco.net** · March 17th 2009, 04:15 AM

do you get a final answer of $18(x^3-3x^2+3x-1)<br />$

**mr fantastic** · March 17th 2009, 04:46 AM

Originally Posted by ben.mahoney@tesco.net

do you get a final answer of $18(x^3-3x^2+3x-1)<br />$

Is that the same as the answer moo gave you ....?

Originally Posted by Moo

Hello,

Since you integrate over the unit square, $x \in [0,1]$ and $y \in [0,1]$

so the integral over y should go from 0 to 1-x.

And this gives something like ${\color{red}3(1-x)^2}$

**ben.mahoney@tesco.net** · March 17th 2009, 11:20 AM

i get $-3(1-x^2)$ but you have to multiply this by the constant (6-6x) which gives my answer?

**Moo** · March 17th 2009, 11:31 AM

Originally Posted by ben.mahoney@tesco.net

i get $-3(1-x^2)$ but you have to multiply this by the constant (6-6x) which gives my answer?

It would help a lot if you show your working. It'd be far easier to spot out your mistake.
I've just made the integration again, and I still found 3(1-x)²... :s
If Mr F didn't say anything, then there is some chance that it is correct

And why do you want to multiply by (6-6x) ? o.O

**ben.mahoney@tesco.net** · March 17th 2009, 11:54 AM

integrate 6[1-x-y] from 0 to 1-x
you can take the (6-6x) out the front of the integral as its just a constant.
this gives (6-6x) integral -6y
integrating -6y gives -3y^2
putting 1-x in makes the -3(x-1)^2 multiplied by the constant (6-6x)

**Moo** · March 17th 2009, 12:04 PM

Originally Posted by ben.mahoney@tesco.net

integrate 6[1-x-y] from 0 to 1-x
you can take the (6-6x) out the front of the integral as its just a constant.
this gives (6-6x) integral -6y
integrating -6y gives -3y^2
putting 1-x in makes the -3(x-1)^2 multiplied by the constant (6-6x)

(6-6x) is not a constant that you simply pull out from the integral... And it doesn't even multiply the integrand !

If a is a constant, then $\int af=a \int f$ , but that's all you can say !

You can pull out 6, nothing else.
$\int_0^{1-x} 6(1-x-y) ~dy=6 \int_0^{1-x} (1-x)-y ~dy=I$
It is true that (1-x) is a constant, but only with respect to y.

You can separate the integral :
$I=6 \int_0^{1-x} (1-x) ~dy-6 \int_0^{1-x} y ~dy$
$I=6(1-x) \int_0^{1-x} 1 ~dy-6 \int_0^{1-x} y ~dy$
$I=6(1-x) \times \left. y \right|_{y=0}^{y=1-x} -6 \times \left. \frac{y^2}{2} \right|_{y=0}^{y=1-x}$

$I=6(1-x) [(1-x)-0]-6 \left[\frac{(1-x)^2}{2}-\frac 02\right]$

...

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